f(x)=ln(x). Find the area underneath the curve f(x) between 1 and 2.

We cannot dirrectly intergrate ln(x), so instead we intergrate 1ln(x) using intergration by parts.
The formula for intergration by parts is: ∫ (udv/dx) dx = uv − ∫ vdu/dx dx .
We let u=ln(x) so that du/dx=1/xWe let dv/dx=1 so that v=x
We put those values into the formula and we get ∫ ln(x) dx = xln(x) - ∫ (x1/x )dx∫ ln(x) dx = xln(x) - ∫1 dx∫ ln(x) dx = xln(x)-x + c
Finding the area under the curve between 1 and 2. ∫²₁ ln(x) dx = [xln(x)-x]²₁ ∫²₁ ln(x) dx = 2ln(2)-2-(1ln(1)-1) ∫²₁ ln(x) dx = 2*ln(2)-1